OCR Salters F331 23/05/13 discussion

learn them then buddy

How many different calculations are there do you reckon, on the whole syllabus?

Do we need to know the solubility rules?


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Do we need to know the solubility rules?

can anyone explain this question to me: June 12 3)d)
it's a calculation question I don't understand

Yes, it's pretty easy once you get your head around it:

The question is asking for the volume of air required for combustion of 60cm³

We know that the ration of isooctane to oxygen is 1 to 12.5 (1:12.5)

So we need to multiply the amount of moles of oxygen by the volume of isooctane (both are 60 times larger, still follow ratio)

12.5 * 60 = 750 (Volume of oxygen in the air mixture)

Because we know that air is 21% oxygen, we can say that in 100 molecules, 21 of them are oxygen. Therefore the amount of moles of oxygen to isooctane:

100 / 21 = 4.7619047619

Ratio =
1 : 4.7619047619

So, there must be this times many more isooctane than oxygen:

750 * 4.7619047619 = 3571.42857143

The question requires this is dm³ therefore we need to divide by 1000

3571.42857143 / 1000 = 3.57dm³

I'm sorry for the late reply. I hope this helped.